Problem: $g(t) = 3t^{2}+4t+2(f(t))$ $h(t) = -2t^{2}-t+f(t)$ $f(t) = -7t^{2}-7t$ $ h(g(0)) = {?} $
Explanation: First, let's solve for the value of the inner function, $g(0)$ . Then we'll know what to plug into the outer function. $g(0) = 3(0^{2})+(4)(0)+2(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = -7(0^{2})+(-7)(0)$ $f(0) = 0$ That means $g(0) = 3(0^{2})+(4)(0)+(2)(0)$ $g(0) = 0$ Now we know that $g(0) = 0$ . Let's solve for $h(g(0))$ , which is $h(0)$ $h(0) = -2(0^{2})-0+f(0)$ To solve for the value of $h$ , we need to solve for the value of $f(0)$ $f(0) = -7(0^{2})+(-7)(0)$ $f(0) = 0$ That means $h(0) = -2(0^{2})-0$ $h(0) = 0$